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julia-pushkina [17]
3 years ago
6

QUESTION 9 of 10: Mutual funds can be:

Mathematics
2 answers:
JulijaS [17]3 years ago
7 0
Maybe A because that’s seem the best
lesya [120]3 years ago
3 0

Answer: d) All of the above

Step-by-step explanation:

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I need help with this please
frez [133]
A = (6 x 7) + 1/2(6 x 3)
= 42 + 9
= 51
answer
51 in^2
5 0
3 years ago
Read 2 more answers
The length of a rectangle field is represented by the expression 14 X minus 3X squared +2 Y. The width of the field is represent
Assoli18 [71]

Answer:

9x+4x^2-5y

Step-by-step explanation:

Hi there!

Length of the field: 14x-3x^2+2y units

Width of the field: 5x-7x^2+7y units

To find how much greater the length of the field is than the width, subtract the width from the length:

14x-3x^2+2y-(5x-7x^2+7y)

Open up the parentheses

= 14x-3x^2+2y-5x+7x^2-7y

Combine like terms

= 14x-5x-3x^2+7x^2+2y-7y\\= 9x+4x^2-5y

Therefore, the length is 9x+4x^2-5y units greater than the width.

I hope this helps!

7 0
3 years ago
Solve by graphing:<br> x^2-6x+9=0
irinina [24]
Hello :
the solution :x²-6x+9=0 is the x- intresept and graph of f(x) =x²-6x+9
the solution is :  x=3

6 0
3 years ago
Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
3 years ago
What kind of solution(s) do you expect for the linear equation:
Kryger [21]
One solution A. Thats the correct answer
7 0
3 years ago
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